Urn balls without replacement, probability on nth position

An urn contains $w$ white and $b$ black balls. $n$ extractions without replacement are made. What are the probabilty of:

• get a black ball on $i-th$ extraction?
• get a black ball on $i-th$ extraction and white on $j-th$ extraction, with $j > i$?

I know that the hypergeometric distribution doesn't care about position, so what is the math behind this problem?

• For the first question, ask yourself, which of the $(w+b)$ balls is/are more or less likely to be the one chosen at the $i-th$ turn? For the second question, you can reason similarly but keep in mind that the two events are not quite independent.
– Ned
Feb 20 '16 at 16:52
• For the second, by symmetry it is the same as the probability of black on the first and white on the second. This is easy. Feb 20 '16 at 17:11
• @Ned I would say $\frac{b}{w+b}$, but I don't really understand why.
– Paul
Feb 20 '16 at 17:39
• yes for the first question, because each of the balls is equally likely to be the $i-th$ one chosen. Thinking about conditional probability and/or the previous choices here just adds extraneous information that isn't needed to answer the question.
– Ned
Feb 20 '16 at 21:23
• @Ned So, if I understand the symmetry the probability to get a black ball on $i−th$ extraction, white on $j−th$ extraction, black on $k−th$ extraction and white on $z−th$ extraction, with $z>k>j>i$ is $\frac{b}{w+b}\frac{w}{w+b-1}\frac{b-1}{w+b-2}\frac{w-1}{w+b-3}$, right?
– Paul
Feb 21 '16 at 15:46

Just imagine them all placed randomly in a row, their position won't change by extraction

Then P(a black ball is in any position) $=\frac{b}{w+b}$,

and P(a white ball is in any position) $= \frac{w}{w+b}$.

This directly gives the answer for the first part

• P(get a black ball on i−th extraction) $=\frac{b}{w+b}$

• For P(get a black ball on i−th extraction and white on j−th extraction), the logic is more subtle, the probabilities of a $B-W$ pair occupying any two positions will be the same, hence the same as $B-W$ occupying positions $1$ and $2$, $=\frac{bw}{(b+w)(b+w-1)}$

• I think I'm closer. The probability to get a black ball on first draw is $\frac{b}{w+b}$. If I get a black ball on first draw then the probability to get a black ball on second draw is $\frac{b-1}{w+b-1}$, but in this case I conditioned the result.
– Paul
Feb 20 '16 at 18:11
• Suppose there is 1 white ball and 1 black ball. Let i=1 and j=2. Clearly the probability of black then white is 1/2 not 1/4 as suggested by the above answer. Feb 20 '16 at 18:17
• My answer to the 2nd part was incorrect. I will amend. Feb 20 '16 at 18:20
• So, using the symmetry stuff provided by André Nicolas the answer to 2nd part is $\frac{b}{w+b}\frac{w}{w+b-1}$?
– Paul
Feb 20 '16 at 18:25
• That's right. In my revision, I have indicated the type of symmetry involved. Feb 20 '16 at 18:44
• Number of all $i$ extractions are $S(i)=\binom{n}{i}*i!$

• Number of cases where to get a black ball on $i$th extraction is $B(i)$

$B(i)=b*\binom{n-1}{i-1}(i-1)!$

probability is $P(i)=\frac{B(i)}{S(i)}$

• Number of cases where to get a black ball on $i$th then white on $j$th is $W(i)$

$W(j)=b*w*\binom{n-2}{j-2}(j-2)!$

probability is $P(j)=\frac{W(j)}{S(j)}$

Simplest way to visualize this (at least for me) is to forget the fact that there are balls of two colors in the urn, and just imagine that each ball in the urn is a 'mixture' of both the colors.

Now, it is easy to see that each draw doesn't change any thing w.r.t. the distribution. That way the probability is independeny of each draw.